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Background information-:

Here is the definitions about what the timestamps mean-:

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Value of original timestamp: 46

Value of receive timestamp: 59

Value of transmit timestamp: 60

Time the packet arrived: 67

Thus Sending time = receive timestamp=original timestamp=59 - 46 = 13 milliseconds

Receiving time =time message arrived at sender-transmit timestamp= 67 - 60 = 7 milliseconds

Round-trip time =sending time+receiving time=13 + 7 = 20 milliseconds

Now say we Consider that receiving node's clock=3+sending node's clock.(I know it is really a basic level math but I am confused here as my initituion isn't working)

The the slide writes-:

Sending time = (56+3) – 46 = 10 + 3 Receiving time = 67 – (57+3) = 10 – 3 RTT = (10 + 3) + (10-3) = 20

Here is where I get confused. I personally believe that it should be-: sending time=(59+3)-46=13+3(as receiving time=3+sending time) receiving time=67+3-60=10

But I am wrong. But I can't comprehend what is written in that slide.

BTW here is the slide-: http://osnet.cs.nchu.edu.tw/powpoint/Computer%20Network93_1/Chapter%209.pdf

It is page number 67/86

I know this might be really a dumb question to ask, but I have been banging my head for hours and unable to find a solution of this. Please guide.

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1 Answer 1

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You are misinterpreting what they are saying (because they are not clear in how they say it). Their intent is that the clock at the receiver is $3$ ms ahead of the clock at the sender. So

  • Receiver timestamp of $59$ occurs at the same time as $56$ on the Sender clock: $59 = 56 + 3$.
  • Receiver timestamp of $60$ occurs that the same time as $57$ on the Sender clock: $60 = 57 + 3$.

So if the Receiver clock were synchonized to the Sender clock, the timestamps on its end would be $56$ and $57$, and the calculation would look like

  • sending time $= 56 - 46 = 10$ ms.
  • receiving time $= 67 - 57 = 10$ ms.
  • RTT $ = 10 + 10 = 20$ ms.

While the actual calculation with unsynchronized clocks is

  • sending time $= 59 - 46 = 13$ ms.
  • receiving time $= 67 - 60 = 7$ ms.
  • RTT $ = 13 + 7 = 20$ ms.

The error in the clock synchronization changes the sending time and receiving time in opposite directions. But when you add them to get the round-trip time, the changes cancel each other, and the round-trip time is not changed. The point is the round-trip time is unaffected by this almost-unavoidable error, while the two one-way times are inaccurate.

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